Question

How do you prove that the limit of `(x^3*y^2)/(x^2+y^2)` as (x,y) approaches (0,0) using the epsilon delta proof?

Answer 1

See explanation

Explanation:

If we have a function `f(x)` of one variable, then:

`lim_(x->c) f(x) = a" "` if and only if:

`AA epsilon > 0 EE delta > 0 : AA x in (c - delta, c + delta)" " abs(f(x) - a) < epsilon`

This can be generalised to a function from `RR^2 -> RR` as follows:

`lim_((x, y) -> (c, d)) f(x, y) = a" "` if and only if:

`AA epsilon > 0 EE delta > 0 : AA x in (c - delta, c + delta) AA y in (d - delta, d + delta)" " abs(f(x, y) - a) < epsilon`

In our example:

`f(x, y) = (x^3y^2)/(x^2+y^2)`

`(c, d) = (0, 0)`

and we will prove that:

`lim_((x, y) -> (0, 0)) = 0`

Given any `epsilon > 0`

Let:

`delta = { (sqrt(2) " if " epsilon > sqrt(2)), (epsilon " if " epsilon <= sqrt(2)) :}`

Let:

`x in (0 - delta, 0 + delta)`

`y in (0 - delta, 0 + delta)`

Then:

`x^2 < delta^2 <= 2`

`y^2 < delta^2 <= 2`

So:

`1/x^2+1/y^2 > 1/2+1/2 = 1`

So:

`1/(1/x^2+1/y^2) < 1`

Then:

If `x = 0` and `y = 0` then `f(x, y)` is undefined. So `(0, 0)` is not part of the domain of `f(x, y)`

If `x = 0` and `y != 0` or `x != 0` and `y = 0` then `f(x, y) = 0` and:

`abs(f(x, y) - 0) = 0 < epsilon`

Otherwise:

`abs(f(x, y) - 0) = abs((x^3y^2)/(x^2+y^2))`

`color(white)(abs(f(x, y) - 0)) = abs(x)(x^2y^2)/(x^2+y^2)`

`color(white)(abs(f(x, y) - 0)) = abs(x)(1)/(1/y^2+1/x^2)`

`color(white)(abs(f(x, y) - 0)) < abs(x) < delta < epsilon`